在数列an中,a1a2a3.....an=n²,求an

a1+a2+a3=7,a1a2a3+8?如果是a1+a2+a3=7,a1a2a3=8的话：a1a2a3=8即a2ˆ3=8,得a2=2,由a1+a2+a3=7,得q=1/2或2

a1a2=4,a1a2a3=9,所以a3=9/4,a1a2a3a4=16,a1a2a3a4a5=25,所以a5=25/16a3+a5=9/4+25/16=61/16

因为a1a2a3=8所以a2/q*a2*a2*q=8a2^3=8,a2=2又a1+a2+a3=7即a2/q+a2+a2*q=71/q+q=5/2=2+1/2所以q=2或1/2即a1=1或4.所以an=

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

a1=a1a2=a1qa3=a1q^2a1(1+q+q^2)=14a1a2a3=a1^3q^3=64a1q=4a1=4/q代入,4(1+q+q^2)=14q整理,得2q^2-5q+2=0(q-2)(2

才2个条件是求不出的,需加多一个条件.如为等差数列,得3a2=-3,a2(a2-d)(a2+d)=8,得a2=-1,d=3或-3,{a1,a2,a3}={-4,-1,2}如为等比数列,得a2/q+a2

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

n-3条,AA3..AAn-1

a1+a1q+a1q^2=7a1^3q^3=8a1q=2a1+2+a1q^2=7a1+a1q^2=5a1=2/q2/q+2/q*q^2=52/q+2q=52+2q^2=5q2q^2-5q+2=0(2q

1a1=2,a2=a1+c=2+c,a3=a2+2c=2+c+2c=2+3c.因a1,a2,a3成公比不为1的等比数列,所以a2^2=a1*a3,即(2+c)^2=2*(2+3c).整理得：c^2-2

a1a3=a2^2a1a2a3＝8a2=2a1＋a2＋a3＝7a1+a3=5a1*a3=4a1=1a3=4q=2a1=4a3=1q=1/2(舍)Sn=a1(1-q^n)/(1-q)=2^n-1

设b1=a1a4a7...a28;b2=a2a5a8...a29;b3=a3a6a9...a30,则有b3=b2*2^10=b1*2^20,所以b2=2^(30/3)=2^10,故b3=2^20,即答

⑴若a1＋a2＋a3=21,a1a2a3=216,设a1=a2/q,a3=a2qa2/q＋a2＋a2q=21a2³=216=6³a2=66/q＋6＋6q=211/q＋q=5/2=1

题为：在数列{a[n]}中,a[1]=2,a[n+1]=a[n]+cn(c是常数),且a[1]、a[2]、a[3]成等比数列,求数列{(a[n]-c)/(n.c^n)}的前n项之和T[n].其中[&n

n+Sn=2an,所以1+s1=2a1=2s1即s1=a1=1且n+1+S(n+1)=2a(n+1)相减得1+a(n+1)=2a(n+1)-2ana(n+1)=2an+1a(n+1)+1=2an+2=

∵a1＝35，a2＝31100∴a2−110a1＝14，a2−12a1＝1100∵数列{an+1−110an}是公比为12的等比数列，首项为a2−110a1＝14∴an+1−110an=14(12)n

这么懒,求a3而已a1=1a1a2=4a1a2a3=9a3=9/4一般an=a1a2a3…an/a1a2a3…a(n-1)=n平方/(n-1)平方=[n/(n-1)]平方

迅速算的话,可以这样算：a1a2a3=a2(a1a3)=a2^3=27,a2=3a1+a2+a3=a2(1+q+1/q)=13,即q+1/q=10/3,得q=3所以an=3^(n-1)

n-3,A1A3,A1A4.A1An-1n-3n-3n(n-3)/2

a1a2a3…an=n^2n=1时a1=1n>1时a1a2a3…an=n^2(1)a1a2a3…a(n-1)=(n-1)^2(2)(1)/(2)=an=n^2/(n-1)^2a3=9/4a5=25/1